Question

The distance of the point $\left(1,3,-7\right)$ from the plane passing through the point $\left(1,-1,1\right),$ having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1},$ is:

• A. $\frac{10}{\sqrt{74}}$
• B. $\frac{20}{\sqrt{74}}$
• C. $\frac{4}{\sqrt{83}}$
• D. $\frac{10}{\sqrt{5}}$

Answer 1

The correct option is C $\frac{4}{\sqrt{83}}$

Let the DR of the normal of the plane be $<a,b,c>$

Since it is perpendicular to $<1,-2,3>$ and $<2,-1,-1>$

$\therefore a-2b+3c=0-\left(1\right)$

$2a-b-c=0-\left(2\right)$

on solving

$\frac{a}{2+3}=\frac{-b}{-1-6}=\frac{c}{-1+4}$

$a=5,b=7,c=3$

$\therefore$ Equation of plane passing through $(1,-1,-1)$

$\Rightarrow 5(x-1)+7(y+1)+3(z-1)=0$

$\Rightarrow 5x-5+7y+7+3z-3=0$

$\Rightarrow 5x+7y+3z-1=0$

Distabce of $(1,3,-7)$ from the plane,

$=\left|\frac{5\left(1\right)+3\left(7\right)+3(-7)-1}{\sqrt{5^2+7^2+3^2}}\right|$

$=\frac{5+21-21-1}{\sqrt{83}}$

$=\frac{4}{\sqrt{83}}$